Answer:
D
Explanation:
The decrease in potential energy is equal to the increase in kinetic energy.
mgh
250 x 9.8 x 30
=73, 500
Equations of motion (EoM) use EoM <span>v2=u2+2ax</span> to establish velocities at positions shown in blue in drawing from EoM v=u+at for final 1 second of flight time, we can say v=u+g(1) <span><span>2gH−−−−√</span>=<span><span>2g1625H</span>−−−−−−√</span>+g</span><span> then, solve for H [in terms of g]
</span>
The slope of the road can be given as the ratio of the change in vertical
distance per unit change in horizontal distance.
- The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.
Reasons:
Width of the truck = 2.4 meters
Height of the truck = 4.0 meters
Height of the center of gravity = 2.2 meters
Required:
The allowable steepness of the slope the truck can be parked without tipping over.
Solution:
Let, <em>C</em> represent the Center of Gravity, CG
At the tipping point, the angle of elevation of the slope = θ
Where;

The steepness of the slope is therefore;

Where;
= Half the width of the truck =
= 1.2 m
= The elevation of the center of gravity above the ground = 2.2 m



The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.
Learn more here:
brainly.com/question/20793607
Answer:
f = 931.1 Hz
Explanation:
Given,
Mass of the wire, m = 0.325 g
Length of the stretch, L = 57.7 cm = 0.577 m
Tension in the wire, T = 650 N
Frequency for the first harmonic = ?
we know,

μ is the mass per unit length
μ = 0.325 x 10⁻³/ 0.577
μ = 0.563 x 10⁻³ Kg/m
now,

v = 1074.49 m/s
The wire is fixed at both ends. Nodes occur at fixed ends.
For First harmonic when there is a node at each end and the longest possible wavelength will have condition
λ=2 L
λ=2 x 0.577 = 1.154 m
we now,
v = f λ


f = 931.1 Hz
The frequency for first harmonic is equal to f = 931.1 Hz
Answer:
Force of <u>1</u><u>0</u><u>0</u><u> </u><u>N</u><u> </u>will be needed.
Explanation:

g is acceleration due to gravity
m is the mass
